# $\sum_{k=1}^{n} \arccos k$

Compute $$\lim_{n \to \infty} \sum_{k=1}^{n} \arccos k$$.
My attempt :the sequence is strictly increasing and if I prove that it is not upper bounded its limit is $$\infty$$. However, I can't prove the last part.

• Are you computing $\arccos 2$ with complex numbers? – J.G. Nov 14 '18 at 16:06
• What is $\arccos(10)$? – ablmf Nov 14 '18 at 16:06

## 2 Answers

$$\cos^{-1}(x)=\frac{\pi}{2}+i\log(\sqrt{1-x^2}+ix)$$ So $$\lim_{x \to \infty} \frac{1}{i}\cos^{-1}(x)=\infty$$

• Isn't arccos multivalued? – N. S. Nov 14 '18 at 16:11
• @N.S. Is it? I think it's up to your definition. Of course, $\cos^{-1}$ is formally not a function but a relation, but you can define an "arccos" function. – Botond Nov 14 '18 at 16:15

Well, $$\arccos(k)$$ is not defined for $$k\ge 1$$ when we consider it as a real function. So the question only makes sense when we consider it as a complex function. Thus $$\cos ^{-1}(k)=\frac{\pi }{2}+i \log \left(\sqrt{1-k^2}+i k\right).$$ Then obviously $$\sum_{k \ge 1} \cos ^{-1}(k)$$ does not converge.